Maths WorkshopMaths problems with full solutions
Exercise 1612Combinatorics › Counting principlesOlympiadAdded 3 Nov 2026

Derangement formula

#derangement#inclusion-exclusion#formula

Statement

Prove that Dn=∑k=0n(−1)k(nk)(n−k)!D_n=\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-k)!.

Hint

Use inclusion-exclusion on the set of fixed points.

Show solutionHide solution

Solution

Setup. Let AiA_i = permutations fixing element ii. We want permutations fixing no element, i.e. A1∪⋯∪An‾\overline{A_1\cup\cdots\cup A_n}.

Count. ∣Ai1∩⋯∩Aik∣=(n−k)!|A_{i_1}\cap\cdots\cap A_{i_k}|=(n-k)! (permutations fixing kk elements arbitrarily).

Inclusion-exclusion.
Dn=∑k=0n(−1)k(nk)(n−k)!=n!∑k=0n(−1)kk!.D_n=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)!=n!\sum_{k=0}^n\frac{(-1)^k}{k!}.

Conclusion. Dn=n!∑k=0n(−1)kk!D_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}.